561_PartUniversity Physics Solution

561_PartUniversity Physics Solution - Thermal Properties of...

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Thermal Properties of Matter 18-19 18.78. IDENTIFY: rms 3 RT v M = . SET UP: 2 3 H 2.02 10 kg/mol M . 2 3 O 32.0 10 kg/mol M . For Earth, 24 5.97 10 kg M and 6 6.38 10 m R . For Jupiter, 27 1.90 10 kg M and 7 6.91 10 m R . For a sphere, 3 4 3 M Vr ρρ π == . The escape speed is escape 2 GM v R = . EXECUTE: (a) Jupiter: 33 rms 3(8.3145J mol K)(140K) (2.02 10 kg mol) 1.31 10 m s v =⋅ × = × . 4 escape 6.06 10 m/s v . rms escape 0.022 vv = . Earth: rms 3(8.3145J mol K)(220K) (2.02 10 kg mol) 1.65 10 m s v × = × . 4 escape 1.12 10 m/s v . rms escape 0.15 = . (b) Escape from Jupiter is not likely for any molecule, while escape from earth is much more probable. (c) 3 rms 3(8.3145J mol K)(200K) (32.0 10 kg mol) 395m s. v × = The radius of the asteroid is 1/3 5 (3 4 ) 4.68 10 m, RM πρ × and the escape speed is escape 2 542m s vG M R . Over time the 2 O molecules would essentially all escape and there can be no such atmosphere. EVALUATE: As Figure 18.23 in the textbook shows, there are some molecules in the velocity distribution that have speeds greater than rms v . But as the speed increases above rms v the number with speeds in that range decreases. 18.79. IDENTIFY: rms 3 kT v m = . The number of molecules in an object of mass m is AA m N nN N M . SET UP: The volume of a sphere of radius r is 3 4 3 = . EXECUTE: (a) 23 14 22 rms 3 3(1.381 10 J K)(300K) 1.24 10 kg. (0.0010m s) kT m v × (b) 14 23 3 A (1.24 10 kg)(6.023 10 molecules mol) (18.0 10 kg mol) Nm NM −− × × × 11 4.16 10 molecules. N (c) The diameter is 1/3 1/3 1/3 14 6 3 3 3 / 3(1.24 10 kg) 2 2 2 2 2.95 10 m 4 4 4 (920 kg/m ) Vm Dr ρ ππ ⎛⎞ × ⎛⎞ ⎛ = = = × ⎜⎟ ⎜⎟ ⎜ ⎝⎠ ⎝ ⎝⎠ which is too small to see. EVALUATE: rms v decreases as m increases. 18.80. IDENTIFY: For a simple harmonic oscillator, cos x At ω = and sin x vA t =− , with / km = . SET UP: The average value of cos(2 ) t over one period is zero, so 1 av av 2 (sin ) (cos ) tt ωω = = . EXECUTE: cos x = , sin x t , 1 av av 2 (cos ) Uk A t = , 2 1 av av 2 (sin ) K mA t = . Using 1 av av 2 (sin ) (cos ) and 2 mk = shows that av av K U = . EVALUATE: In general, at any given instant of time UK . It is only the values averaged over one period that are equal. 18.81. IDENTIFY: The equipartition principle says that each atom has an average kinetic energy of 1 2 kT for each degree of freedom. There is an equal average potential energy. SET UP: The atoms in a three-dimensional solid have three degrees of freedom and the atoms in a two- dimensional solid have two degrees of freedom. EXECUTE: (a) In the same manner that Eq.(18.28) was obtained, the heat capacity of the two-dimensional solid would be 2 16.6 J/mol K R . (b) The heat capacity would behave qualitatively like those in Figure 18.21 in the textbook, and the heat capacity would decrease with decreasing temperature. EVALUATE: At very low temperatures the equipartition theorem doesn’t apply. Most of the atoms remain in their lowest energy states because the next higher energy level is not accessible.
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18-20 Chapter 18 18.82. IDENTIFY: The equipartition principle says that each molecule has average kinetic energy of 1 2 kT for each degree of freedom. 2 2(/ 2 ) Im L = , where L is the distance between the two atoms in the molecule. 2 1 rot 2 K I ω = .
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561_PartUniversity Physics Solution - Thermal Properties of...

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